Question

A rectangle has a width of x centimeters and a perimeter of 8x centimeters a square has sides of length 1/4 that of the length of the rectangle.
A. Find the length of the rectangle
B. Find the perimeter of the square
C. Find how many cm greater the rectangle's perimeter if x=4
D. Find how many square cm greater the rectangle's area is than the square's area if x=4

Answer 1

Answer:

Part A) The length of rectangle is $3x\ cm$

Part B) The perimeter of the square is $3x\ cm$

Part C) $20\ cm$

Part D) $39\ cm^{2}$

Step-by-step explanation:

Part A) Find the length of the rectangle

we know that

The perimeter of rectangle is equal to

$P=2(L+W)$

we have

$P=8x\ cm$

$W=x\ cm$

substitute and solve for L

$8x=2(L+x)$

$4x=(L+x)$

$L=4x-x=3x\ cm$

Part B) Find the perimeter of the square

we know that

The perimeter of a square is

$P=4b$

we have that

$b=(1/4)L$

substitute the value of L

$b=(1/4)3x=(3/4)x\ cm$

Find the perimeter of the square

$P=4(3/4)x=3x\ cm$

Part C) Find how many cm greater the rectangle's perimeter than the square's perimeter if x=4

Find the value of rectangle's perimeter

$P=8x\ cm$ ------> $P=8(4)=32\ cm$

Find the value of square's perimeter

$P=3x\ cm$ ------> $P=3(4)=12\ cm$

Find the difference

$32\ cm-12\ cm=20\ cm$

Part D) Find how many square cm greater the rectangle's area is than the square's area if x=4

Find the value of rectangle's area

$A=(3x)(x)=3x^{2}\ cm^{2}$ ------> $A=3(4^{2})=48\ cm^{2}$

Find the value of square's area

$A=((3/4)x)^{2}\ cm^{2}$ ------> $A=((3/4)(4))^{2}=9\ cm^{2}$

Find the difference

$48\ cm^{2}-9\ cm^{2}=39\ cm^{2}$